Question: Solve for $k$. $\dfrac k{2}+\dfrac12=3 $
Let's subtract and then multiply to get $k$ by itself. $\begin{aligned} \dfrac k{2}+\dfrac12&=3 \\ \\ \dfrac{k}{2}+\dfrac12{-\dfrac12} &=3{-\dfrac12}~~~~~{\text{subtract }\dfrac12} \text{ from each side}\\ \\ \dfrac k2+\cancel{\dfrac12} {{-}\cancel{{\dfrac12}}}&= 3{ -\dfrac12}\\ \\ \dfrac k{2}&= 3{-\dfrac12}\end{aligned}$ $\begin{aligned}\dfrac k2&= \dfrac52 \\ \\ \dfrac k{2}\cdot{{2}}&= {\dfrac52}\cdot{{2}} ~~~~~~~\text{multiply each side by } {2} \text{ to get } k \text{ by itself }\\ \\ \dfrac k{\cancel{2}}\cdot{\cancel{{2}}} &={\dfrac52}\cdot{{2}} \\ \\ k&= {\dfrac52}\cdot{{2}} \end{aligned}$ The answer: $k={5}$ Let's check to make sure. $\begin{aligned} \dfrac k{2}+\dfrac12&=3 \\\\ \dfrac{{5}}{2}+\dfrac12&\stackrel{?}{=} 3 \\\\ \dfrac62&\stackrel{?}{=} 3 \\\\ 3&= 3 ~~~~~~~~~~\text{Yes!} \end{aligned}$